so because they used the word "prove," i am hesitant to tell you to just plug in 5 for x in the function and show that you get 11 -- which you can do since the function is a polynomial and hence is continuous everywhere.

however, a proof is as follows.

Definition:
A limit is formally defined as follows: Let be a function defined on an open interval containing (except possibly at ) and let be a real number. means that
for each real there exists a real such that for all where , .So, to prove Lim x->5(x^2-3x+1) = 11, we must show that for every e > 0 there exists a d > 0 such that for all x in dom(f) and 0 < |x - 5|< d implies |x^2-3x+1 - 11|=|x^2-3x-10|< e